Thermodynamics Lecture SeriesVapor Cycle – Ideal Rankine Cycle Overcoming Impracticalities of...
Transcript of Thermodynamics Lecture SeriesVapor Cycle – Ideal Rankine Cycle Overcoming Impracticalities of...
Thermodynamics Lecture Series
Ideal Ideal Rankine Rankine Cycle Cycle ––The The Practical CyclePractical Cycle
Applied Sciences Education Research Group (ASERG)
Faculty of Applied SciencesUniversiti Teknologi MARA
email: [email protected]://www5.uitm.edu.my/faculties/fsg/drjj1.html
Steam Power PlantExample: A steam power cycle.Example: A steam power cycle.
SteamTurbine Mechanical Energy
to Generator
Heat Exchanger
Cooling Water
Pump
Fuel
Air
CombustionProducts
System Boundaryfor ThermodynamicAnalysis
System Boundaryfor ThermodynamicAnalysis
Second LawSecond LawHigh T Res., TH
Furnace
qin = qH
Low T Res., TLWater from river
qout = qL
Working fluid:Water Purpose:
Produce work,Wout, ωout
Steam Power Plant ωnet,out
An Energy-Flow diagram for a SPP
Second Law – Dream EngineSecond Law – Dream Engine
Carnot CycleP - ν diagram for a Carnot (ideal) power plantP, kPa
ν, m3/kgqout
qin
2
34
1
What is the maximum performance of real engines if it can never achieve 100%??
in
out,net
qnputi equiredroutput desired ω
η ==
revin
outinrev q
−=η
Carnot Principles• For heat engines in contact with the same hot
and cold reservoir P1: η1 = η2 = η3 (Equality)P2: ηreal < ηrev (Inequality)
Second Law – Will a Process HappenSecond Law – Will a Process Happen
revreal ηη ≤
;(K) (K)
H
L
revH
L
TT
=
(K) (K) 11
H
L
revH
Lrev T
Tqq
−=
−=η
Consequence
Processes satisfying Carnot Principles obeys the Second Law of Thermodynamics
Clausius Inequality :• Sum of Q/T in a cyclic process must be zero
for reversible processes and negative for real processes
∫ ≤KkJ ,0
TQδ
Second Law – Will a Process HappenSecond Law – Will a Process Happen
∫ = ,0TQδ
∫ < ,0TQδ
∫ •≤
KkgkJ ,0
Tqδ
reversiblereal
∫ > ,0TQδ impossible
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Isolated systemsFIGURE 6-6The entropy change of an isolated system is the sum of the entropy changes of its components, and is never less than zero.
6-3
Entropy – Quantifying DisorderEntropy – Quantifying DisorderIncrease of Entropy Principle – closed system
The entropy of an isolated (closed and adiabatic) system undergoing any process, will always increase.
Surrounding
System
0≥∆+∆=+=∆ surrsysgenheatisolated SSSSS
)ss(mS 12sys −=∆
( )surr
surroutinsurr T
QQS
−=∆
For pure substance:
( )surrTQ
geninnetssmS ,)( 12 +−=
and
Then
Entropy – Quantifying DisorderEntropy – Quantifying DisorderEntropy Balance – for any general system
For any system undergoing any process,
Energy must be conserved (Ein – Eout = ∆Esys)
Mass must be conserved (min – mout = ∆msys)
Entropy will always be generated except for reversible processes
Entropy balance is (Sin – Sout + SSgengen = ∆Ssys)
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Entropy Transfer
FIGURE 6-61Mechanisms of entropy transfer for a general system.
6-18
Entropy – Quantifying DisorderEntropy – Quantifying DisorderEntropy Balance –Steady-flow device
outinoutin WWQQ••••
−+− kW ,mminout
−
=
••ϑϑ
0SSSS sysgenoutin ==+−•••
∆ inoutgen SSS,So•••
−=
Then:
inmassheat
outmassheatgen SSSSS
+−
+=
•••••
inletin
in
exitout
outgen sm
TQsm
TQS
−−
+= ∑∑
••
••
•
Entropy – Quantifying DisorderEntropy – Quantifying DisorderEntropy Balance –Steady-flow device
outinoutin WWQQ••••
−+− ( ) kW ,m inletexit ϑϑ −=•
( ) kW ,000 34 hhmW out −=−+−••
Assume adiabatic, ∆kemass = 0,∆pemass = 0
where•••
== mmm exitinlet
( )K
kWssmS gen ,00 34 −+−=••
EntropyBalance K
kWsmsmTQ
TQS
in
in
out
outgen ,3344
••••
•
−+−=In,3
Out
Turbine:
Entropy – Quantifying DisorderEntropy – Quantifying DisorderEntropy Balance –Steady-flow device
outinoutin WWQQ••••
−+− kW ,mminletexit
∑∑
−
=
••ϑϑ
Mixing Chamber:
KkW,smsmsm
TQ
TQS 112233
in
in
out
outgen
•••••
•−−+−=
exitinlet mm••
=
kW,hmhmhmWWQQ 112233outinoutin •••••••
−−=−+−
1
3
2where
Vapor Cycle Vapor Cycle Steam Power Plant
External combustionFuel (qH) from nuclear reactors, natural gas, charcoal Working fluid is H2OCheap, easily available & high enthalpy of vaporization hfg
Cycle is closed thermodynamic cycleAlternates between liquid and gas phaseCan Can Carnot Carnot cycle be used for representing real SPP??cycle be used for representing real SPP??Aim: To decrease ratio of TL/TH
Vapor Cycle – Carnot CycleVapor Cycle – Carnot CycleEfficiency of a Carnot Cycle SPP
55.0273374
273151TT1
H
Lrev =
++
−=−=η
627.0273500
273151TT1
H
Lrev =
++
−=−=η
Vapor Cycle –Carnot CycleVapor Cycle –Carnot CycleImpracticalities of Carnot Cycle
Isothermal expansion: TH limited to only Tcrit for H2O.High moisture at turbine exitNot economical to design pump to work in 2-phase (end of Isothermal compression)No assurance can get same xfor every cycle (end ofIsothermal compression)
s3 = s4s1 = s2
qin = qHT, °C
Tcrit
TH
TL
qout = qL
s, kJ/kg°K
Vapor Cycle – Alternate Carnot CycleVapor Cycle – Alternate Carnot CycleImpracticalities of Alternate Carnot Cycle
s3 = s4s1 = s2
qin = qHT, °C
Tcrit
TH
TL
qout = qL
s, kJ/kg°K
Still ProblematicIsothermal expansion but at
variable pressurePump to very high pressure
Can the boiler sustain the high P?
Vapor Cycle – Ideal Rankine CycleVapor Cycle – Ideal Rankine Cycle
Overcoming Impracticalities of Carnot Cycle
SuperheatSuperheat the H2O at a constant pressure (isobaric expansion)Can easily achieve desired TH higher than Tcrit.reduces moisture content at turbine exit
Remove all excess heat at condenserPhase is sat. liquid at condenser exit, hence need only a pump to increase pressure Quality is zero for every cycle at condenser exit(pump inlet)
Vapor Cycle – Ideal Rankine CycleVapor Cycle – Ideal Rankine Cycle
Pum
p
Boiler Turbine
Condenser
qin = qH
ωout
Low T Res., TLWater from river
qout = qLqin - qout = ωout - ωin
qin - qout = ωnet,out
High T Res., THFurnace
Working fluid:Water
ωin
A Schematic diagram for a Steam Power Plant
Vapor Cycle – Ideal Rankine CycleVapor Cycle – Ideal Rankine CycleT- s diagram for an Ideal Rankine Cycle
T, °C
s, kJ/kg°K
1
2
Tcrit
TH
TL= Tsat@P4
Tsat@P2
s3 = s4s1 = s2
qin = qH
4
3
PHPL
ωin
ωout
pump
qout = qL
condenser
turbineboiler
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FIGURE 9-2The simple ideal Rankinecycle.
9-2
Vapor Cycle – Ideal Rankine CycleVapor Cycle – Ideal Rankine Cycle
BoilerIn,2 Out,3
qin = qH
Energy AnalysisAssume ∆ke =0, ∆pe =0 for the moving mass, kJ/kg
qin – qout+ ωin – ωout = θout – θin, kJ/kg
qin – 0 + 0 – 0 = hexit – hinlet, kJ/kg
Qin = m(h3 – h2), kJqin = h3 – h2, kJ/kg
( ) kWhhmQ in ,23 −=••
Vapor Cycle – Ideal Rankine CycleVapor Cycle – Ideal Rankine CycleEnergy Analysis Assume ∆ke =0, ∆pe =0 for
the moving mass, kJ/kg
qin – qout+ ωin – ωout = θout – θin, kJ/kg
0 – qout + 0 – 0 = hexit – hinlet - qout = h1 – h4,
CondenserOut,1 In,4
qout = qL
( ) kWhhmQ out ,14 −=••
So, qout = h4 – h1, kJ/kg
Qout = m(h4 – h1), kJ
Vapor Cycle – Ideal Rankine CycleVapor Cycle – Ideal Rankine CycleEnergy Analysis
ωout
In,3
Out,4
Turbineqin – qout+ ωin – ωout = θout – θin, kJ/kg
0 – 0 + 0– ωout = hexit – hinlet, kJ/kg
- ωout = h4 – h3, kJ/kg So, ωout = h3 – h4, kJ/kg
Wout = m(h3 – h4), kJ Assume ∆ke =0, ∆pe =0 for the moving mass, kJ/kg( ) kWhhmW out ,43 −=
••
Vapor Cycle – Ideal Rankine CycleVapor Cycle – Ideal Rankine CycleEnergy Analysis
Pum
pωin
Out,2
In,1
∫∫∫ +=+=2
1
2
1
2
1in,pump dP0dPPd νννω
( ) 1212in,pump hhPP −=−=νω
1P@f12 ννν =≅
So, Win = m(h2 – h1), kJ
( ) kWhhmW in ,12 −=••
qin – qout+ ωin – ωout = θout – θin, kJ/kg
0 – 0 + ωin – 0= hexit – hinlet, kJ/kg
ωin = h2 – h1, kJ/kgFor reversible pumps
where
Vapor Cycle – Ideal Rankine CycleVapor Cycle – Ideal Rankine CycleEnergy Analysis
Efficiency
( )23
1423
in
outin
in
out,net
hhhhhh
qqq
q −−−−
=−
==ω
η
( )23
1243
in
inout
in
out,net
hhhhhh
qq −−−−
=−
==ωωω
η
23
1243
hhhhhh
−+−−
==η
Vapor Cycle – Ideal Rankine CycleVapor Cycle – Ideal Rankine CycleT- s diagram for an Ideal Rankine Cycle
T, °C
s, kJ/kg°K
1
2
Tcrit
TH
TL= Tsat@P4
Tsat@P2
s3 = s4s1 = s2
qin = qH
4
3
PHPL
ωin
ωout
pump
qout = qL
condenser
turbineboiler
s1 = sf@P1 h1 = hf@P1
s3 = s@P3,T3
s4 = [sf +xsfg]@P4 = s3
h3 = h@P3,T3
h4 = [hf +xhfg]@P4
4P@fg
4P@f3
sss
x−
=
Note that P1 = P4
1P@f12 ννν =≅h2 = h1 +ν2(P2 – P1); where
Vapor Cycle – Ideal Rankine CycleVapor Cycle – Ideal Rankine CycleEnergy AnalysisIncreasing Efficiency
Must increase ωnet,out = qin – qout
Increase area under process cycleDecrease condenser pressure; P1=P4
Pmin > Psat@Tcooling+10 deg C
Superheat T3 limited to metullargical strength of boiler
Increase boiler pressure; P2=P3
Will decrease quality (an increase in moisture). Minimum x is 89.6%.
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Lowering Condenser PressureFIGURE 9-6The effect of lowering the condenser pressure on the ideal Rankinecycle.
9-4
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Superheating SteamFIGURE 9-7The effect of superheating the steam to higher temperatures on the ideal Rankine cycle.
9-5
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9-6
FIGURE 9-8The effect of increasing the boiler pressure on the ideal Rankinecycle.
Increasing Boiler Pressure
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FIGURE 9-10T-s diagrams of the three cycles discussed in Example 9–3.
9-8
Vapor Cycle – Reheat Rankine CycleVapor Cycle – Reheat Rankine CyclePu
mp
Boiler High
P turbine
Condenser
High T Reservoir, TH
qin = qH
qout = qL
Low
P turbine1
23
4
5
6
qreheat
ωout,1
ωin
ωout,2
Low T Reservoir, TL
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FIGURE 9-11The ideal reheatRankine cycle.
9-9
Vapor Cycle – Reheat Rankine CycleVapor Cycle – Reheat Rankine CycleReheating increases η and reduces moisture in turbine
TL= Tsat@P1
ωin
s5 = s6s1 = s2
Tcrit
TH
Tsat@P4
Tsat@P3
s3 = s4
qout = h6-h1
ωout, II
P4 = P5
P6 = P1
61
5
4
qreheat = h5-h4
qprimary = h3-h2 ωoutP3
3
2
T, °C
s, kJ/kg°K
Vapor Cycle – Reheat Rankine CycleVapor Cycle – Reheat Rankine CycleEnergy Analysis
q in = qprimary + qreheat = h3 - h2 + h5 - h4 qout = h6-h1
ωnet,out = ωout,1 + ωout,2 - ωin = h3 - h4 + h5 - h6 – h2 + h1
( )4523
164523
in
outin
in
out,net
hhhhhhhhhh
qqq
q −+−−−−+−
=−
==ω
η
4523
12654321,
hhhhhhhhhh
qq in
inoutout
in
outnet
−+−+−−+−
=−+
==ωωωω
η
Vapor Cycle – Reheat Rankine CycleVapor Cycle – Reheat Rankine CycleEnergy Analysis
s6 = [sf +xsfg]@P6. Use x = 0.896 and s5 = s6where
h6 = [hf +xhfg]@P6
Knowing s5 and T5, P5 needs to be estimated (usually approximately a quarter of P3 to ensure x is around 89%. On the property table, choose P5 so that the entropy is lower than s5 above. Then can find h5 = h@P5,T5.
Vapor Cycle – Reheat Rankine CycleVapor Cycle – Reheat Rankine CycleEnergy Analysis
where s1 = sf@P1 h1 = hf@P1
s3 = s@P3,T3 = s4. h3 = h@P3,T3
1P@f12 ννν =≅h2 = h1 +ν2(P2 – P1); where
P5 = P4.
From P4 and s4, lookup for h4 in the table. If not found, then do interpolation.